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3 years ago

Which of the following is an example of an even population distribution?

1 answer:

8 0

Even population distribution refers to a type of population distribution in which the arrangement is done in such a way that the distance between neighboring individual is maximized and uniform. This type of population distribution is usually found on the farm land, where the space in which the crops are planted have been carefully measured out. Thus an example of even population distribution is corn planted in a field.

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The graph represents the change in that occurs when food is cooked over a charcoal grill. Which statement correctly explains the

MArishka [77]

Option C is the answer

<h3 /><h3>Small Explanation:-</h3>

The reactants are charcoal that is unlit + oxygen and the products are the burnt charcoal + energy.

(Explanation with formula and reason attached. Check it.)

$> Benjemin360$

6 0

2 years ago

g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th

kati45 [8]

The ball has height y and velocity v at time t according to

y = v₀ t - 1/2 g t ²

and

v = v₀ - g t

where v₀ is its initial speed and g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time t such that

-2.72 m = v₀ t - 1/2 g t ²

-7.94 m/s = v₀ - g t

Solve for t in the second equation:

t = (v₀ + 7.94 m/s)/g

Substitute this into the first equation and solve for v₀ :

-2.72 m = v₀ (v₀ + 7.94 m/s) /g - 1/2 g ((v₀ + 7.94 m/s)/g)²

-2.72 m = v₀²/g + (7.94 m/s) v₀/g - 1/2 (v₀ + 7.94 m/s)²/g

2 (-2.72 m) g = 2v₀² + 2 (7.94 m/s) v₀ - (v₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2v₀² + (15.9 m/s) v₀ - (v₀² + (15.9 m/s) v₀ + 63.0 m²/s²)

-53.3 m²/s² = v₀² - 63.0 m²/s²

v₀² = 9.73 m²/s²

v₀ = 3.12 m/s

3 0

2 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when

yarga [219]

Q1: At the highest point

3 0

3 years ago

A battery charger is connected to a dead battery and delivers a current of 8.9 A for 4.7 hours, keeping the voltage across the b

adell [148]

Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

Voltage, V = 12V

Electrical energy is given as:

E = I*V*t

Where I = Current

V = Voltage/Potential differenxe

t = time in seconds.

E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

3 0

3 years ago

Read 2 more answers

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago